Chapter 8.2, Problem 13E

To determine

To find the total shaded area .

Answer to Problem 13E

The total shaded area is, 16 squared units.

Explanation of Solution

Concept used: Area between curves.

If f and g are continuous with $f\left(x\right)\ge g\left(x\right)$ throughout [a,b], then the area between

The curves $y=f\left(x\right)\text{ and }y=g\left(x\right)$ from a to b is the integral of $\left[f-g\right]$ from ato b.

  $\begin{array}{l}A={\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]}dx\\ \end{array}$

Given information:

The given function is, $y=-x^2+3x\text{ , }y=2x^3-x^2-5x$

  

Calculation: The total shaded area is obtained as,

The total area between the curves from -2 to 2.

  $\begin{array}{l}f\left(x\right)=-x^2+3x\\ f\left(x\right)=2x^3-x^2-5x\end{array}$

Now, area bounded by these region is:

  $\begin{array}{l}\text{ A}={\displaystyle {\int }_a^b\left[f\left(t\right)-g\left(t\right)\right]}dt\\ ={\displaystyle {\int }_{-2}^0\left(2x^3-x^2-5x\right)-\left(-x^2+3x\right)dx}+{\displaystyle {\int }_0^2\left(-x^2+3x\right)-\left(2x^3-x^2-5x\right)dx}\\ ={\displaystyle {\int }_{-2}^0\left(2x^3-8x\right)dx}+{\displaystyle {\int }_0^2-2x^3+8xdx}\\ ={\left[\frac{1}{2}{x}^4-4x^2\right]}_{-2}^0+{\left[-\frac{1}{2}{x}^4+4x^2\right]}_0^2\\ =\left[\frac{1}{2}{\left(0\right)}^4-4{\left(0\right)}^2\right]-\left[\frac{1}{2}{\left(-2\right)}^4-4{\left(-2\right)}^2\right]+\left[\frac{1}{2}{\left(2\right)}^4-4{\left(2\right)}^2\right]-\left[\frac{1}{2}{\left(0\right)}^4-4{\left(0\right)}^2\right]\\ =0-\left(8-16\right)+\left(-8+16\right)\\ =16\end{array}$

Hence, The ottal shaded area is, 16 squared units.